1. Mi-Jung Choi mjchoi@kangwon.ac.kr Dept. of Computer and Science Silberschatz, Galvin and Gagne ©2006 Operating System Principles Chapter 6: Process Synchronization

2. - 2 - Operating System Fall 2009 Ch06 - Process Synchronization Chapter 6: Process Synchronization  Background → atomic operation  The Critical-Section Problem  Peterson’s Solution  Synchronization Hardware  Semaphores  Classic Problems of Synchronization  Monitors

3. - 3 - Operating System Fall 2009 Ch06 - Process Synchronization Background  Concurrent access to shared data may result in data inconsistency  Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes  Suppose that we wanted to provide a solution to the consumer-producer problem that fills all the buffers.  We can do so by having an integer counter that keeps track of the number of items in the buffer.  Initially, counter is set to 0.  It is incremented by the producer after it produces a new item.  It is decremented by the consumer after it consumes a item.

4. - 4 - Operating System Fall 2009 Ch06 - Process Synchronization Shared data among producer and consumer #define BUFFER_SIZE 10 typedef struct { int content; } item; item buffer[BUFFER_SIZE]; int in = 0;// initial state int out = 0;// empty int counter = 0;

5. - 5 - Operating System Fall 2009 Ch06 - Process Synchronization Producer while (TRUE) { // produce an item and put in nextProduced while (counter == BUFFER_SIZE)// is buffer full? ; // do nothing buffer [in] = nextProduced; in = (in + 1) % BUFFER_SIZE; counter ++; }

6. - 6 - Operating System Fall 2009 Ch06 - Process Synchronization Consumer while (TRUE) { while (counter == 0)// is buffer empty? ; // do nothing nextConsumed = buffer[out]; out = (out + 1) % BUFFER_SIZE; counter --; // consume the item in nextConsumed }

7. - 7 - Operating System Fall 2009 Ch06 - Process Synchronization Race Condition  Although the producer and consumer routines are correct separately, they may not function correctly when executed concurrently.  Suppose  that the value of the variable counter is currently 5 and  that the producer and consumer execute the statements “counter++” and “counter--” concurrently.  Following the execution of these two statements, the value of the variable counter may be 4, 5, or 6.  Why?

8. - 8 - Operating System Fall 2009 Ch06 - Process Synchronization Race Condition  count++ could be implemented as register1 = counter register1 = register1 + 1 counter = register1  count-- could be implemented as register2 = counter register2 = register2 - 1 counter = register2  Consider this execution interleaving with “count = 5” initially: S0: producer execute register1 = counter{register1 = 5} S1: producer execute register1 = register1 + 1 {register1 = 6} S2: consumer execute register2 = counter {register2 = 5} S3: consumer execute register2 = register2 - 1 {register2 = 4} S4: producer execute counter = register1 {counter = 6 } S5: consumer execute counter = register2 {counter = 4} counter = 4

9. - 9 - Operating System Fall 2009 Ch06 - Process Synchronization E - Box S - Box 1. data 3. execution 2. operand 4. result Ex. --------------------------- E-box S-box --------------------------- CPU Memory Computer Disk ---------------------------- “counter ++ ” is not ATOMIC Separation of Execution Box & Storage Box

10. - 10 - Operating System Fall 2009 Ch06 - Process Synchronization Race Condition counter++, counter-- are not ATOMIC Producer E - Box S - Box counter counter++ Consumer E - Box counter--

11. - 11 - Operating System Fall 2009 Ch06 - Process Synchronization Example of a Race Condition Memory CPU1 CPU2 P1 P2 X == 2 X = X + 1; X = X – 1; Load X, R1 Inc R1 Store X, R1 Load X, R2 Dec R2 Store X, R2 Interleaved execution?

12. - 12 - Operating System Fall 2009 Ch06 - Process Synchronization Race Condition  We would arrive at this incorrect state  because we allowed both processes to manipulate the variable counter concurrently.  Race Condition  Several processes access and manipulate the same data concurrently and the outcome of the execution depends on the particular order in which the access takes place  To prevent the race condition  We need to ensure that only one process at a time can manipulate the variable counter  We require that the processes be synchronized in some way

13. - 13 - Operating System Fall 2009 Ch06 - Process Synchronization Critical-Section (CS) Problem  A system consisting of n processes {P 0, P 1, …, P n-1 }  Each process has a segment of code, called a critical section, where the process may change common variables, updating a table, writing a file, and so forth.  The system requires that no two processes are executing in their critical section at the same time.  Solution to the critical-section problem  To design a protocol that the processes can use to cooperate.  Each process must request permission to enter its CS.  The section of code implementing this request is the entry section.  The CS may be followed by an exit section  The remaining code is the remainder section

14. - 14 - Operating System Fall 2009 Ch06 - Process Synchronization Critical-Section Problem  The general structure of a typical process P i Do { entry section critical section exit section remainder section } while ( TRUE );

15. - 15 - Operating System Fall 2009 Ch06 - Process Synchronization Solution to Critical-Section Problem  A solution to the critical-section problem must satisfy the following three requirements: 1.Mutual Exclusion - If process P i is executing in its critical section, then no other processes can be executing in their critical sections 2.Progress - If no process is executing in its critical section and there exist some processes that wish to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely – deadlock-free condition 3.Bounded Waiting - A bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted – starvation-free condition Assume that each process executes at a nonzero speed No assumption concerning relative speed of the n processes

16. - 16 - Operating System Fall 2009 Ch06 - Process Synchronization Peterson’s Solution for critical section problem  Two process {P i, P j } solution  Software-based solution  Assume that the LOAD and STORE instructions are atomic; that is, cannot be interrupted.  The two processes share two variables:  int turn;  Boolean flag[2]  The variable turn indicates whose turn it is to enter the critical section.  The flag array is used to indicate if a process is ready to enter the critical section. flag[i] = true implies that process P i is ready!

17. - 17 - Operating System Fall 2009 Ch06 - Process Synchronization Algorithm for Process P i do { flag[i] = TRUE; turn = j; while ( flag[j] && turn == j); CRITICAL SECTION flag[i] = FALSE; REMAINDER SECTION } while (TRUE); // entry section // exit section

18. - 18 - Operating System Fall 2009 Ch06 - Process Synchronization Peterson’s Solution satisfies 3 conditions  Mutual Exclusion  Each P i enters its critical section only if either flag[j]==false or turn==i.  Each P i enters its critical section with flag[i]==true  Both P i and P j cannot enter their critical section at the same time.  Progress  P i can be stuck only if either flag[j]==true and turn==j.  Bounded Waiting  P i will enter the critical section after at most one entry by P j.

19. - 19 - Operating System Fall 2009 Ch06 - Process Synchronization What is the problem in the P i ? do { while (turn != i); CRITICAL SECTION turn = j; REMAINDER SECTION } while (TRUE); do { flag[i] = true; while (flag[j]); CRITICAL SECTION flag[i] = false; REMAINDER SECTION } while (TRUE); satisfies Mutual Exclusion does not satisfy Progress Bounded Waiting satisfies Mutual Exclusion does not satisfy Progress Bounded Waiting

20. - 20 - Operating System Fall 2009 Ch06 - Process Synchronization Dekker’s Algorithm – two-process solution do { flag[i] = TRUE; while (flag[j] ) { if ( turn == j ) { flag[i] = FALSE; while ( turn == j) ; flag[i] = TRUE; } CRITICAL SECTION turn = j; flag[i] = FALSE; REMAINDER SECTION } while (TRUE);

21. - 21 - Operating System Fall 2009 Ch06 - Process Synchronization Synchronization Hardware  Many systems provide hardware support for critical section code  Uni-processors – could disable interrupts  Currently running code would execute without preemption  Generally too inefficient on multiprocessor systems  Operating systems using this are not broadly scalable  Modern machines provide special atomic hardware instructions  Atomic = non-interruptible  Test memory word and set value: TestAndSet()  Swap contents of two memory words: Swap()

22. - 22 - Operating System Fall 2009 Ch06 - Process Synchronization TestAndSet() Instruction boolean TestAndSet (boolean *target) { boolean rv = *target; *target = TRUE; return rv: }  Definition:  This Instruction is atomic  This Instruction is provided by hardware.

23. - 23 - Operating System Fall 2009 Ch06 - Process Synchronization Solution using TestAndSet()  Shared Boolean variable lock., initialized to false.  Solution for Mutual Exclusion : do { while ( TestAndSet (&lock ) ) ; // do nothing CRITICAL SECTION lock = FALSE; REMAINDER SECTION } while ( TRUE); busy waiting

24. - 24 - Operating System Fall 2009 Ch06 - Process Synchronization Swap() Instruction void Swap (boolean *a, boolean *b) { boolean temp = *a; *a = *b; *b = temp: }  Definition:  This Instruction is atomic  This Instruction is provided by hardware.

25. - 25 - Operating System Fall 2009 Ch06 - Process Synchronization Solution using Swap()  Shared Boolean variable lock initialized to FALSE;  Each process has a local Boolean variable key.  Solution for Mutual Exclusion: do { key = TRUE; while ( key == TRUE) Swap (&lock, &key ); CRITICAL SECTION lock = FALSE; REMAINDER SECTION } while ( TRUE); busy waiting

26. - 26 - Operating System Fall 2009 Ch06 - Process Synchronization Solution using TestAndSet()  Shared Boolean variable waiting[n] and lock initialized to FALSE;  Each process has a local Boolean variable key.  Solution for Bounded Waiting: do { waiting [i] = TRUE; key = TRUE; while ( waiting [i] && key ) key = TestAndSet (&lock); waiting [i] = FALSE; CRITICAL SECTION j = (i+1)%n; while ( (j != i) && !waiting[j] ) j=(j+1)%n; if( j == i) lock = FALSE; else waiting [i] = FALSE; REMAINDER SECTION } while ( TRUE); busy waiting

27. - 27 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore  Semaphore Alphabet

28. - 28 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore  The various hardware-based solutions to the critical section problem  TestAndSet(), Swap()  complicated for application programmers to use  To overcome this difficulty  Semaphore may be used.  Semaphore is …  Synchronization tool that does not require busy waiting  Semaphore S – integer variable  Two standard operations to modify S: wait() and signal()  Originally called P() for wait() and V() for signal()  The modification of the semaphore is atomic.  less complicated to use.

29. - 29 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore  Definition for wait(S);  Definition for signal(S);  All the modification to the semaphore is atomic.  When one process modifies the semaphore value, no other process can simultaneously modify that same semaphore value. wait (S) { while ( S <= 0 ) ; // no-op S--; } signal (S) { S++; } busy waiting

30. - 30 - Operating System Fall 2009 Ch06 - Process Synchronization Usage of Semaphore  Counting semaphore  Integer value can range over an unrestricted domain  Ex. 0.. 10  Binary semaphore  Integer value can range only between 0 and 1  Which is refer to as mutex locks 1.Binary semaphore for solving the critical-section problem for multiple processes  n processes share a semaphore, mutex.  mutex is initialized to 1  The structure of a process P i do { wait (mutex); CRITICAL SECTION signal (mutex); REMAINDER SECTION } while (TRUE);

31. - 31 - Operating System Fall 2009 Ch06 - Process Synchronization 1. binary semaphore

32. - 32 - Operating System Fall 2009 Ch06 - Process Synchronization 1. binary semaphore sem_wait() is same to wait(); sem_post() is same to signal();

33. - 33 - Operating System Fall 2009 Ch06 - Process Synchronization 1. binary semaphore  Does previous code satisfy three requirements of CS problem.  Mutual exclusion  Progress  Bounded waiting  The third requirement is not guaranteed as a default.  It usually depends on the implementation of wait() function.  Ex, Linux sem_wait() does not guarantee the bounded waiting req.

34. - 34 - Operating System Fall 2009 Ch06 - Process Synchronization 1. binary semaphore

35. - 35 - Operating System Fall 2009 Ch06 - Process Synchronization 1. binary semaphore

36. - 36 - Operating System Fall 2009 Ch06 - Process Synchronization Usage of Semaphore 2.Counting semaphore used to control access to a given resource consisting of a finite number of instances  Semaphore is initialized to the number of resources available  To use a resource, a process performs wait()  To release a resource, a process perform signal()  When the semaphore is 0, all resources are being used. 3.Counting semaphore to solve various synchronization problems.  Ex. Two concurrently running processes P1, P2  P1 with a statement S1, P2 with a statement S2  S2 is executed only after S1 has completed  How to solve this using semaphore?

37. - 37 - Operating System Fall 2009 Ch06 - Process Synchronization Usage of Semaphore  Solution of #3 in previous page  Initialization  P1 structure  P2 structure Semaphore synch = 0; S1; signal (synch); wait (synch); S2;

38. - 38 - Operating System Fall 2009 Ch06 - Process Synchronization 3. counting semaphore

39. - 39 - Operating System Fall 2009 Ch06 - Process Synchronization 3. counting semaphore

40. - 40 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore Implementation with Busy waiting  must guarantee that no two processes can execute wait () and signal () on the same semaphore at the same time  Thus, implementation becomes the critical section problem where the wait and signal code are placed in the critical section.  Previous codes have busy waiting in critical section implementation  While a process in critical section, others must loop continuously in wait code.  called a spinlock  Disadvantage  waists CPU cycle during wait();  Sometimes it is useful:  No context switch involved in the wait(), signal()  Implementation code is short  Little busy waiting if critical section rarely occupied  However, applications may spend lots of time in critical sections and therefore this is not a good solution.

41. - 41 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore Implementation with no Busy waiting  To overcome the busy waiting problem, two operations are used  Block() – place the process invoking the operation on the appropriate waiting queue.  Wakeup() – remove one of processes in the waiting queue and place it in the ready queue.  When a semaphore value is not positive on executing wait(), the process blocks itself instead of busy waiting.  signal() operation wakes up a waiting process.  To implement semaphores under this definition, we define a semaphore as a record like  With each semaphore there is an associated waiting queue. typedef struct { int value; // semaphore value struct process *list; // pointer to the PCB list of waiting processes }

42. - 42 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore Implementation with no Busy waiting  How to implement the waiting queue of a semaphore?  linked list in the semaphore  contains PCBs of waiting processes.  The negative value means the number of waiting processes  The positive value means the number of available resources  The list can use any queuing strategy.  To satisfy the bounded waiting requirements the queue can be implemented with FIFO queue.  Two pointer variables which indicate head and tails of the PCB list. value:-3 list PCB

43. - 43 - Operating System Fall 2009 Ch06 - Process Synchronization Semaphore Implementation with no Busy waiting  Implementation of wait()  Implementation of signal() wait ( semaphore *S) { S->value --; if ( S->value < 0 ) { add this process to S->list; // put the PCB into waiting queue block(); // go to waiting state from running state } signal ( semaphore *S) { S->value ++; if ( S->value <= 0 ) { remove a process P from S->list; // select a process from waiting queue wakeup(); // put the process into ready queue }

44. - 44 - Operating System Fall 2009 Ch06 - Process Synchronization Deadlock and Starvation  Deadlock – two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes  The use of a semaphore with a waiting queue may result in a deadlock.  Let S and Q be two semaphores initialized to 1  Starvation – indefinite blocking. A process may never be removed from the semaphore queue in which it is suspended.  The implementation of a semaphore with a waiting queue may result in a starvation.  When the queue is in LIFO (last-in, first-out) order. P 0 wait (S); wait (Q); … signal (S); signal (Q); P 1 wait (Q); wait (S); … signal (Q); signal (S);

45. - 45 - Operating System Fall 2009 Ch06 - Process Synchronization Classical Problems of Synchronization  Bounded-Buffer Problem  Readers and Writers Problem  Dining-Philosophers Problem

46. - 46 - Operating System Fall 2009 Ch06 - Process Synchronization Bounded-Buffer Problem  More than two producers  produce one item, store it in the buffer, and continue.  More than two consumers  consume a item from buffer, and continue.  The buffer contains at most N items.  Solution with semaphore  Semaphore mutex initialized to the value 1  Semaphore full initialized to the value 0  Semaphore empty initialized to the value N.

47. - 47 - Operating System Fall 2009 Ch06 - Process Synchronization Bounded Buffer Problem  The structure of the producer process do { // produce an item wait (empty); // check buffer is full or not wait (mutex); // enter critical section // add the item to the buffer // critical section signal (mutex); // exit critical section signal (full); // one item is produced } while (TRUE);

48. - 48 - Operating System Fall 2009 Ch06 - Process Synchronization Bounded Buffer Problem  The structure of the consumer process do { wait (full); // check buffer is empty or not wait (mutex); // enter critical section // add the item to the buffer // critical section signal (mutex); // exit critical section signal (empty); // one item is produced // consume the item } while (TRUE);

49. - 49 - Operating System Fall 2009 Ch06 - Process Synchronization Readers-Writers Problem  A data set is shared among a number of concurrent processes  Readers – only read the data set; they do not perform any updates  Writers – can both read and write.  Problem – allow multiple readers to read at the same time. Only one single writer can access the shared data at the same time.  Shared Data  Data set  Semaphore mutex initialized to 1.  Semaphore wrt initialized to 1.  Integer readcount initialized to 0.

50. - 50 - Operating System Fall 2009 Ch06 - Process Synchronization Readers-Writers Problem  The structure of a writer process do { wait (wrt); // enter a critical section // writing is performed // critical section signal (wrt); // exit critical section } while (TRUE);

51. - 51 - Operating System Fall 2009 Ch06 - Process Synchronization Readers-Writers Problem  The structure of a reader process do { wait (mutex); readcount ++; if ( readcount == 1 ) wait(wrt); // when it is the first reading signal(mutex) // reading is performed // more than two reader can read // concurrently wait(mutex); readcount --; if ( readcount == 0 ) signal(wrt); // when it is the last reading signal (mutex); } while (TRUE);

52. - 52 - Operating System Fall 2009 Ch06 - Process Synchronization Readers-Writers Problem  Starvation is the problem…  The writer may starve.  No readers will be kept waiting unless a writer has already obtained permission to use the shared object.  Another variation of Readers-Writers Problem  Once a writer is ready, that the writer performs its write as soon as possible  If a writer is waiting to access the object, no new readers may start reading  => In this case, the readers may starve.  What is the starvation free solution for Readers-Writers Problem?

53. - 53 - Operating System Fall 2009 Ch06 - Process Synchronization Dining-Philosophers Problem  Five philosophers who spend their lives thinking and eating.  They share a circular table surrounded by five chairs.  In the center of the table is a bowl of rice.  File single chopsticks on the table.  When a philosophers think, she does not interact with neighbors.  From time to time, she gets hungry and try to pick up the two chopsticks that are closest to her.  She picks up only one chopsticks at a time.  She cannot pick up the chopstick that is already in the hand of a neighbor.  When she gets two chopsticks, she eats without releasing them.  When she finished eating, she put down both chopsticks and think again.  How to implement this with concurrent processors in a deadlock-free and starvation-free manner?

54. - 54 - Operating System Fall 2009 Ch06 - Process Synchronization Dining-Philosophers Problem  It is a large class of concurrency-control problems.  It is a simple representation of the need to allocate several resources among several processes in a deadlock-free and starvation-free manner.  Simple Solution …  Shared data  Bowl of rice (data set)  Semaphore chopstick [5] initialized to 1

55. - 55 - Operating System Fall 2009 Ch06 - Process Synchronization Dining-Philosophers Problem (Cont.)  The structure of Philosopher i: do { wait ( chopstick [i] ); wait ( chopstick [ (i+1)%5 ] ); … // eat … signal ( chopstick [i] ); signal ( chopstick [ (i+1)%5 ] ); … // think … } while (TRUE);

56. - 56 - Operating System Fall 2009 Ch06 - Process Synchronization Dining-Philosophers Problem (Cont.)  This solution guarantees that  No two neighbors are eating simultaneously  However, this solution creates deadlock.  Suppose all five philosophers become hungry simultaneously and each grabs her left chopsticks.  All chopsticks are taken.  When each philosophers tries to grab her right chopstick, she delays forever  Possible deadlock free solutions  Allow at most four philosophers to be sitting together  Allow a philosopher to pick up her chopsticks only if both chopsticks are available  Use an asymmetric solution; an odd one picks up left chopstick first, an even one picks up right chopstick first.

57. - 57 - Operating System Fall 2009 Ch06 - Process Synchronization Problems with Semaphores  Be careful not to misuse the semaphore.  Examples of misuse:  mutex is a semaphore which is initialized to 1  signal (mutex) …. wait (mutex)  results in a violation of the mutual exclusion requirement  wait (mutex) … wait (mutex)  results in deadlock  Omitting of wait (mutex) or signal (mutex) (or both)  Either mutual exclusion is violated or a deadlock will occur  To deal with such errors, the monitor may be used.

58. - 58 - Operating System Fall 2009 Ch06 - Process Synchronization Monitors  A high-level abstraction that provides a convenient and effective mechanism for process synchronization  Monitor is defined as an abstract data type like class in object-oriented language (C++, Java)  A monitor instance is shared among multiple processes.  Only one process may be active within the monitor at a time monitor monitor-name { shared variable declarations procedure P 1 (…) { …. } … procedure P n (…) {……} Initialization code ( ….) { … } } Private Data Public Methods Initialization Method

59. - 59 - Operating System Fall 2009 Ch06 - Process Synchronization Schematic view of a Monitor  Shared data  Operations  Initialization code  The shared data can be accessed by operations  Only one process is active within a monitor at a time  Programmer don’t need to code this synchronization.

60. - 60 - Operating System Fall 2009 Ch06 - Process Synchronization Condition Variables  Basic monitor is not sufficiently powerful for modeling some synchronization schemes.  To solve, condition variables are introduced.  condition x, y;  A programmer can define one or more variables of type condition  Two operations on a condition variable:  x.wait () – a process that invokes the operation is suspended.  x.signal () – resumes one of processes (if any) that invoked x.wait ()  x.signal() resumes exactly one suspended process.  If no process is suspended, then nothing happens.

61. - 61 - Operating System Fall 2009 Ch06 - Process Synchronization Monitor with Condition Variables

62. - 62 - Operating System Fall 2009 Ch06 - Process Synchronization Condition Variables – issues  Suppose that, when the x.signal() is invoked by a process P, there is a suspended process Q with condition x.  If the suspended process Q is allowed to resume its execution, the signaling process P must wait.  Otherwise both Q and P would be active simultaneously.  Two possibilities:  Signal and Wait: P either waits until Q leaves the monitor or waits for another condition  Signal and Continue: Q either waits until P leaves the monitor or waits for another condition.

63. - 63 - Operating System Fall 2009 Ch06 - Process Synchronization Solution to Dining Philosophers monitor DP { enum { THINKING; HUNGRY, EATING) state [5] ; condition self [5]; void pickup (int i) { state[i] = HUNGRY; test(i); if (state[i] != EATING) self [i].wait(); } void putdown (int i) { state[i] = THINKING; // test left and right neighbors test((i + 4) % 5); test((i + 1) % 5); } Three stats of a philosopher Ph_i delays herself when she is hungry and unable to pick both chopsticks Ph_i invokes the operation pickup(i) before she eat. Ph_i invokes the operation putdown(i) after she eat.

64. - 64 - Operating System Fall 2009 Ch06 - Process Synchronization Solution to Dining Philosophers (cont) void test (int i) { if ( (state[(i + 4) % 5] != EATING) && (state[i] == HUNGRY) && (state[(i + 1) % 5] != EATING) ) { state[i] = EATING ; self[i].signal () ; } initialization_code() { for (int i = 0; i < 5; i++) state[i] = THINKING; } Ph_i invokes the test(i) to pickup chopsticks. Ph_i invokes the test(i+1) and test((i+1)%5) to release her chopsticks. When two neighbors are not eating, she changes its state to EATING and invoke her. Initialize all the Ph with THINKING state.

65. - 65 - Operating System Fall 2009 Ch06 - Process Synchronization Solution to Dining Philosophers (cont) do { thinking dp.pickup( i ) eat. dp.putdown ( i ); } while (TRUE);  A philosopher_i must invoke the operation pickup() and putdown() in the following sequence.

66. - 66 - Operating System Fall 2009 Ch06 - Process Synchronization self[1] self[2] pickup() putdown() test() condition variable queue entry queue shared data operations initialization_code() pickup() putdown() Solution to Dining Philosophers (cont)

67. - 67 - Operating System Fall 2009 Ch06 - Process Synchronization Monitor Implementation using Semaphores  Variables semaphore mutex; // (initially = 1) semaphore next; // (initially = 0) int next-count = 0;  Each external procedure F will be replaced by wait(mutex); … body of F; … if (next-count > 0) signal(next); else signal(mutex);  Mutual exclusion within a monitor is ensured.

68. - 68 - Operating System Fall 2009 Ch06 - Process Synchronization Monitor Implementation using Semaphores  For each condition variable x, we have: semaphore x-sem; // (initially = 0) int x-count = 0;  The operation x.wait can be implemented as: x-count++; if (next-count > 0) signal(next); else signal(mutex); wait(x-sem); x-count--;

69. - 69 - Operating System Fall 2009 Ch06 - Process Synchronization Monitor Implementation using Semaphores  The operation x.signal can be implemented as: if (x-count > 0) { next-count++; signal(x-sem); wait(next); next-count--; }

70. - 70 - Operating System Fall 2009 Ch06 - Process Synchronization Monitor Implementation  Conditional-wait operation: x.wait(c);  c – integer expression evaluated when the wait operation is executed.  value of c (a priority number) stored with the name of the process that is suspended.  when x.signal is executed, process with smallest associated priority number is resumed next.  Check two conditions to establish correctness of system:  User processes must always make their calls on the monitor in a correct sequence.  Must ensure that an uncooperative process does not ignore the mutual-exclusion gateway provided by the monitor, and try to access the shared resource directly, without using the access protocols.

71. - 71 - Operating System Fall 2009 Ch06 - Process Synchronization Summary  Cooperating processes that share data have critical section of code each of which is in use by only one process at a time.  Solutions for critical section problem  Software-based solution: Peterson’s algorithm, Bakery algorithm  Hardware-based solution: Testandset(), Swap()  Three requirements for critical section problem  Mutual Exclusion, Progress, Bounded Wating  The main disadvantage of these solution is that they all require busy waiting. Semaphore overcome this difficulty.  Semaphore can be used to solve various synchronization problems  The bounded-buffer, The readers-writers, The dining-philosophers problem problems  Those are examples of a large class of concurrency-control problems.  The solution should be deadlock-free ad starvation-free.  Monitors provides the synchronization mechanism for sharing abstract data types.  Condition variables provides a method by which a monitor procedure can block its execution until it is signaled to continue.

72. Mi-Jung Choi mjchoi@kangwon.ac.kr Dept. of Computer and Science Silberschatz, Galvin and Gagne ©2006 Operating System Principles End of Chapter 6